Every idempotent matrix over a field is diagonalizable

Let A be an n \times n matrix over a field F. Show that if A^2 = A, then A is similar to a diagonal matrix whose diagonal entries are either 0 or 1.

(Compare to this previous exercise.)

Short Proof that I Thought Of Seconds after Publishing the Long Proof:

The minimal polynomial of A divides x^2-x = x(x-1). By Corollary 25 in D&F, A is diagonalizable. Moreover, the elementary divisors of A are all either x or x-1, so the diagonal entries of the Jordan canonical form of A are either 0 or 1.

Long Proof that Missed the Point of the Whole Chapter but to which I am Too Attached to Delete Wholesale:

Let A be such a matrix, and let J be the Jordan canonical form of A; say P^{-1}AP = J. Now J^2 = (P^{-1}AP)^2 = P^{-1}APP^{-1}AP = P^{-1}A^2P = P^{-1}AP = J, so that J is also idempotent. Note also that J is upper triangular, with all entries above the first superdiagonal equal to 0. That is, if J = [t_{i,j}], then t_{i,j} = 0 if j \neq i or j \neq i+i and t_{i,j} \in \{0,1\} if j = i+1. (The diagonal entries of J are unknown at this time.) Since J^2 = J, we have \sum_k t_{i,k}t_{k,j} = t_{i,j} for each pair (i,j).

Fix i. Now t_{i,i} = \sum_k t_{i,k}t_{k,i} = t_{i,i}t_{i,i} + \sum_{k \neq i} t_{i,k}t_{k,i} = t_{i,i}t_{i,i}. That is, t_{i,i}^2 = t_{i,i}, so that t_{i,i} is a root of x^2-x = x(x-1) = 0. So t_{i,i} (i.e. an arbitrary diagonal entry) is either 1 or 0.

Now fix i and let j = i+1. Now t_{i,j} = \sum_k t_{i,k}t_{k,i+1} = t_{i,i}t_{i,i+1} + t_{i,i+1}t_{i+1,i+1} + \sum_{k \neq i,i+1} t_{i,k}t_{k,i+1} = t_{i,i}t_{i,i+1} + t_{i,i+1}t_{i+1,i+1}. Note that if t_{i,i+1} = 1, then we must have t_{i,i} = t_{i+1,i+1}, since these entries are in the same Jordan block of J. Now 1 = 2t_{i,i}, and we have 1 = 0, a contradiction. Thus $latex t_{i,i+1} = 0 for all appropriate i. That is, J is diagonal.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: